Question: The equilibrium concentrations of the species in the reaction $\mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C}+\mathrm{D}$ are $2,3,10$ and $6 \mathrm{~mol} \mathrm{~L}^{-1}$, respectively at $300 \mathrm{~K} . \Delta \mathrm{G}^0$ for the reaction is $(\mathrm{R}=2 \mathrm{cal} / \mathrm{mol} \mathrm{K})$
A) $-137.26 \mathrm{cal}$
B) $-1381.80 \mathrm{cal}$
C) $-13.73 \mathrm{cal}$
D) $1372.60 \mathrm{cal}$
Answer: $-1381.80 \mathrm{cal}$
Solution:
$$ \begin{aligned} \mathrm{A}+\mathrm{B} & \rightleftharpoons \mathrm{C}+\mathrm{D} \ {[A] } & =2 \mathrm{~mol} \mathrm{~L}^{-1} \ {[B] } & =3 \mathrm{~mol} \mathrm{~L}^{-1} \ {[C] } & =10 \mathrm{~mol} \mathrm{~L}^{-1} \ {[D] } & =6 \mathrm{~mol} \mathrm{~L}^{-1} \ \Delta \mathrm{G}^0 & =-2.303 \mathrm{RT} \log \mathrm{K}_{\mathrm{eq}} \ & =-2.303 \mathrm{RT} \log \frac{[\mathrm{C}][\mathrm{D}]}{[\mathrm{A}][\mathrm{B}]} \ & =-2.303 \times 2 \times 300 \times \log \frac{10 \times 6}{2 \times 3} \ & =-2.303 \times 2 \times 300 \times \log 10 \ & =-1381.8 \mathrm{cal} \ & \end{aligned} $$
